Solution:

The vector eq. of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\overrightarrow{\mathrm{b}}$ is $\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}$
Given: the line passes through $(1,2,-4)$

Therefore,
$\overrightarrow{\mathrm{a}}=1 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}$
Also given that, line is parallel to both planes.
Therefore it can be said that the line is perpendicular to normal of both planes.
That is $\vec{b}$ is perpendicular to normal of both planes.

It is known that
$\vec{a} \times \vec{b}$ is perpendicular to both $\vec{a} \& \vec{b}$
Therefore, $\overrightarrow{\mathrm{b}}$ is cross product of normal of planes
$\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}$ and $\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}$

$\begin{array}{l}
\text { Required Normal }=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & -16 & 7 \\
3 & 8 & -5
\end{array}\right| \\
=\hat{i}[(-16)(-5)-8(7)]-\hat{j}[3(-5)-3(7)]+\hat{k}[3(8)-3(-16)] \\
=\hat{i}[80-56]-\hat{j}[-15-21]+\hat{k}[24+48] \\
=24 \hat{i}+36 \hat{j}+72 \hat{k}
\end{array}$

So,
$\overrightarrow{\mathrm{b}}=24 \hat{\mathrm{i}}+36 \hat{\mathrm{j}}+72 \hat{\mathrm{k}}$
Substituting the value of $\vec{a} \& \vec{b}$ in the formula, we obtain

$\begin{aligned}
\vec{r} =\vec{a}+\lambda \vec{b} \\
=(1 \hat{i}+2 \hat{j}-4 \hat{k})+\lambda(24 \hat{i}+36 \hat{j}+72 \hat{k}) \\
=(1 \hat{i}+2 \hat{j}-4 \hat{k})+12 \lambda(2 \hat{i}+3 \hat{j}+6 \hat{k}) \\
=(1 \hat{i}+2 \hat{j}-4 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})
\end{aligned}$

As a result, the equation of the line is
$\vec{r}=(1 \hat{\boldsymbol{i}}+2 \hat{\boldsymbol{j}}-4 \hat{\boldsymbol{k}})+\lambda(2 \hat{\boldsymbol{i}}+3 \hat{\boldsymbol{j}}+6 \hat{\boldsymbol{k}})$